Two interior angles $A$ and $B$ of pentagon $ABCDE$ are $60^{\circ}$ and $85^{\circ}$. Two of the remaining angles, $C$ and $D$, are equal and the fifth angle $E$ is $15^{\circ}$ more than twice $C$. Find the measure of the largest angle.
Solution: The sum of the angle measures in a polygon with $n$ sides is $180(n-2)$ degrees.  So, the sum of the pentagon's angles is $180(5-2) = 540$ degrees.

Let $\angle C$ and $\angle D$ each have measure $x$, so $\angle E = 2x + 15^\circ$.  Therefore, we must have \[60^\circ + 85^\circ + x + x+ 2x + 15^\circ = 540^\circ.\] Simplifying the left side gives $4x + 160^\circ = 540^\circ$, so $4x = 380^\circ$ and $x = 95^\circ$.  This means the largest angle has measure $2x + 15^\circ = 190^\circ + 15^\circ = \boxed{205^\circ}$.